Ask & Answer: India's Largest Education Community

No, MGKVP Varanasi provides merit-based admissions to its BBA programme. Candidates are also required to fulfil the preset eligibility criteria. Eligible candidates can apply for the university-level counselling which is based on the merit list released by the university on its official website. The seats are allocated to the candidates who manage to make it to the merit-list. Hence, direct admission to the BBA programme of MGVP may not be possible.

A coherent light source in Young's Double Slit Experiment illuminates two closely spaced slits, and produces two overlapping light waves. The interference of these waves constructively or destructively based on their phase difference lead to a pattern of bright and dark fringes on a screen, When the path difference is an integral multiple of the wavelength, it is constructive interference (bright fringes) and when the path difference is an odd multiple of half the wavelength, it is destructive interference (dark fringes). Through observable interference patterns, Young's Double Slit Experiment, shows the wave nature of light.

13.6 Volume of the room, V = 25.0 $m^3$

Temperature of the room, T = 27 $?$ = 300 K

Pressure of the room, P = 1 atm = 1 $\times$ 1.013 $\times {10}^5$ Pa

The ideal gas equation relating to pressure (P), volume (V) and absolute temperature (T) can be written as

PV = $k_B$ NT, where $k_B$ is Boltzmann constant = 1.38 $\times {10}^{-23}$ $m^{2}kg{s}^{-2}{K}^{-1}$

N is the number of air molecules in the room.

N = $\frac{PV}{k_{B}T}$ = $\frac{1\mathrm{}\times 1.013\mathrm{}\times {10}^5\times 25}{1.38\mathrm{}\times {10}^{-23}\times 300}$ = 6.117 $\times {10}^{26}$

Yes, the Ramrao Adik Institute of Technology has begun the M.Tech admission process for the academic year 2025. Candidates willing to take admission can fill out the application form by visiting the official website of the institute. However, the institute has not yet released the deadline to apply. Students seeking admission are advised to apply in a timely manner.

4.20

(a) The position is given by $\stackrel{?}{r}$ = 3. 0t ? − 2.0t 2 ? + 4.0 ?

The velocity v is given by $\stackrel{?}{v}$ = $\frac{d\stackrel{?}{r}}{dt}$ = $\frac{d}{dt}$ (3.0t ? − 2.0t 2 ? + 4.0 ?)

$\stackrel{?}{v}$ = 3.0 ? – 4.0t?

Acceleration $\stackrel{?}{a}$ = $\frac{d\stackrel{?}{v}}{dt}$ = $\frac{d}{dt}$ (3.0 ? – 4.0t?) = – 4.0?

(b) At t = 2.0s

$\stackrel{?}{v}$ = 3.0 ? – 4.0t?

The magnitude of velocity is given by $\left|\stackrel{?}{v}\right|$ = (3.02 +- 8.02)0.5 = 8.54 m/s

Direction, $\theta$ = ${\tan }^{-1}?\frac{v_y}{v_x}$ = ${\tan }^{-1}?\frac{8}{3}$ = 69.44 $°$

At the UG level MGKVP Varanasi provides a BBA programme lasting three years. students looking for admission to the programme must fulfil basic eligibility criteria specified by university. To be eligible for admission students must have passed Class 12/equivalent examination from a recognised board. Additionally they must present necessary documents supporting their eligibility at time of admission.

After a BBA from DSMNRU, one has many diversified business administration careers to choose from. One could be an investment banker, a research assistant, a financial manager, an operations analyst, and so on. Some of the most sought-after BBA career options for DSMNRU students are outlined below:

4.19

(a) The net acceleration of a particle in circular motion is always directed along the radius of the circle towards the centre (centripetal force), in the case of a circular uniform motion. Hence the statement is False.

(b) The centrifugal force direction is always along the tangent, hence the statement is True.

(c) In uniform circular motion, the direction of the acceleration vector points is always toward the centre of the circle. The average of these vectors over one cycle is a null vector. Hence the statement is True.

4.18 The radius of the loop, r = 1 km = 1000 m

Speed, v = 900 km/h = 250 m/s

Centripetal acceleration, AC = $\frac{v^2}{r}$ = 250 $*250$ / 1000 m/s² = 62.5 m/s² = $\frac{62.5}{9.817}$ g = 6.37g

4.17 Given, the length of the string, l = 80 cm, No. of revolution = 14, Time taken = 25 s

We know Frequency, v = $\frac{No.ofrevolution}{timetaken}$ = $\frac{14}{25}$ Hz

Angular frequency $\omega$ = 2?v = 2 $\times$ $\frac{22}{7}$ $\times \frac{14}{25}$ rad/s = 3.52 rad/s

Centripetal acceleration ac = ${\omega }^2$ r = ${3.52}^2\times \frac{80}{100}$ m/s2 = 9.91 m/s2

The direction of acceleration is towards the centre.

4.16 We know for a projectile motion, horizontal range is given by

R = $\frac{u^2}{g}sin2\theta$ , where R is the horizontal range, u is the velocity and $\theta$ is the angle of projectile

So $\frac{u^2}{g}$ = 100/sin 2 $\theta$

The cricketer will only be able to throw the ball to the maximum horizontal distance when the angle of projection $\theta$ = 45 $°$ , $\frac{u^2}{g}$ = 100 …….(1)

The ball will achieve max height when it is thrown vertically upward. For such motion, final velocity v = 0

From the equation $v^2$ - $u^2$ =2gH, where acceleration a = -g, we get

0 - $u^2$ = -2gH, H = $\frac{u^2}{2g}$ = $\frac{100}{2}$ = 50m

4.15 The speed of the ball, u = 40 m/s

Ceiling height, h = 25m

Since the ball is thrown, it will follow the path of a projectile. For projectile motion, the maximum height for a body projected at an angle $\theta$ is given by

h= $\frac{u^2{sin}^2\theta }{2g}$ , sin2 $\theta$ = (25 $\times 2\times 9.807)/(40\times 40)$ , $\theta =33.61°$

Horizontal range is given by,=

R = $\frac{u^2}{g}sin2\theta$ = ((40 $\times 40$ ) $\times \sin ?2\times 33.61°$ )/9.81 = 150.38m

In case there are errors in the IBPS RRB admit card, candidates must report to the examination authority for correction. Candidates will then get the corrected admit card.

GATE is a national-level entrance exam accepted by many MTech-offering colleges, including Ramrao Adik Institute of Technology. For the year 2025, the GATE exam has already been conducted. The exam was held in February 2025. Students who couldn't appear for the exam can apply with DYPRAIT-PG scores.